Second-order differential subordinations on a class of analytic functions defined by Rafid-operator
The purpose of the present paper is to introduce a new class of analytic functions by using the Rafid-integral operator and obtain some subordination results.
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2018
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Ukrains’kyi Matematychnyi Zhurnal| _version_ | 1860507384817909760 |
|---|---|
| author | Akgül, A. Акгюл, А. |
| author_facet | Akgül, A. Акгюл, А. |
| author_sort | Akgül, A. |
| baseUrl_str | https://umj.imath.kiev.ua/index.php/umj/oai |
| collection | OJS |
| datestamp_date | 2019-12-05T09:19:33Z |
| description | The purpose of the present paper is to introduce a new class of analytic functions by using the Rafid-integral operator and
obtain some subordination results. |
| first_indexed | 2026-03-24T02:08:28Z |
| format | Article |
| fulltext |
UDC 517.5
A. Akgül (Mersin Univ., Turkey)
SECOND-ORDER DIFFERENTIAL SUBORDINATIONS
ON A CLASS OF ANALYTIC FUNCTIONS DEFINED BY RAFID-OPERATOR
ДИФЕРЕНЦIАЛЬНI ПIДПОРЯДКУВАННЯ ДРУГОГО ПОРЯДКУ
НА КЛАСI АНАЛIТИЧНИХ ФУНКЦIЙ,
ЩО ВИЗНАЧЕНI ОПЕРАТОРОМ РАФIДА
The purpose of the present paper is to introduce a new class of analytic functions by using the Rafid-integral operator and
obtain some subordination results.
За допомогою оператора Рафiда введено новий клас аналiтичних функцiй. Отримано деякi результати щодо пiдпо-
рядкування.
1. Introduction. Let \BbbC be complex plane and \BbbU = \{ z \in \BbbC : | z| < 1\} = \BbbU \setminus \{ 0\} , open unit disc in
\BbbC . Let H(\BbbU ) be the class of functions analytic in \BbbU . For n \in N = \{ 1, 2, 3, . . .\} and a \in \BbbC , let
H[a, n] the subclass of H(\BbbU ) consisting of the form
f(z) = z + anz
n + an+1z
n+1 + . . .
with H0 \equiv H[0, 1], H \equiv H[1, 1]. Let An be the class of all analytic functions of the form
f(z) = z +
\infty \sum
k=n+1
akz
k (1.1)
in the open unit disk \BbbU with A1 = A. A function f \in H(\BbbU ) is univalent if it is one to one in \BbbU . Let
S denote the subclass of A consisting of functions univalent in \BbbU . If a function f \in A maps \BbbU onto
a convex domain and f is univalent, then f is called a convex function. Let
K =
\biggl\{
f \in A : \Re
\biggl\{
1 +
zf \prime \prime (z)
f \prime (z)
\biggr\}
> 0, z \in \BbbU
\biggr\}
denote the class of all convex functions defined in \BbbU and normalized by f(0) = 0, f \prime (0) = 1.
Let f and F be members of H(\BbbU ). The function f is said to be subordinate to F, if there exists
a Schwartz function w analytic in \BbbU with
w(0) = 0 and | w(z)| < 1, z \in \BbbU ,
such that
f(z) = F (w(z)).
In such a case we write
f(z) \prec F (z) or f \prec F.
c\bigcirc A. AKGÜL, 2018
ISSN 1027-3190. Укр. мат. журн., 2018, т. 70, № 5 587
588 A. AKGÜL
Furthermore, if the function F is univalent in \BbbU , then we have the following equivalence [6, 12]:
f(z) \prec F (z) \Leftarrow \Rightarrow f(0) = F (0) and f(U) \subset F (U).
The method of differential subordinations (also known the admissible functions method) was
first introduced by Miller and Mocanu in 1978 [10] and the theory started to develop in 1981 [11].
All the details captured in a book by Miller and Mocanu in 2000 [6]. Recent years, many authors
investigated properties of differential subordinations (see [1, 3, 4, 8] and others).
Let \Psi : \BbbC 3 \times \BbbU - \rightarrow \BbbC and h be univalent in \BbbU . If p is analytic in \BbbU and and satisfies the
second-order differential subordination
\Psi
\bigl(
p(z), zp\prime (z), zp\prime \prime (z); z
\bigr)
\prec h(z), (1.2)
then p is called a solution of the differential subordination. The univalent function q is called a
dominant of the solution of the differential subordination or more simply dominant, if p \prec q for all
p satisfying (1.2). A dominant q1 satisfying q1 \prec q for all dominants q of (1.2), is said to be the
best dominant of (1.2).
Recently, Athsan and Buti [6] introduced Rafid-operator of f \in R for 0 \leq \mu < 1, 0 \leq \theta < 1 is
denoted by R\theta
\mu and defined as follows:
R\theta
\mu f(z) =
1
(1 - \mu )\theta +1\Gamma (\theta + 1)
\infty \int
0
t\theta - 1e
-
\Bigl(
t
1 - \mu
\Bigr)
f (zt) dt. (1.3)
Thus, if f \in A is of the form (1.1), we can obtain from (1.3) that
R\theta
\mu f(z) = z +
\infty \sum
k=2
L(k, \mu , \theta )akz
k, (1.4)
where L(k, \mu , \theta ) = (1 - \mu )k - 1\Gamma (k + \theta )
\Gamma (\theta + 1)
.
Using the equation (1.4), it is easily seen that
R\theta
\mu
\bigl(
zf \prime (z)
\bigr)
= z
\Bigl(
R\theta
\mu f(z)
\Bigr) \prime
.
We adopte methods of [5] and introduce a new class by using Rafid-operator R\theta
\mu :
Definition 1.1. Let \Re \mu ,\theta (\beta ) be the class of functions f \in A satisfying
\Re
\biggl\{ \Bigl(
R\theta
\mu f(z)
\Bigr) \prime \biggr\}
> \beta ,
where z \in \BbbU , 0 \leq \beta < 1 and R\theta
\mu is the Rafid-operator.
In order to prove our main results we should need the following lemmas:
Lemma 1.1 [5]. Let h be convex function with h(0) = a and let \gamma \in \BbbC \ast := \BbbC \setminus \{ 0\} be a
complex number with \Re \{ \gamma \} \geq 0. If p \in H[a, n] and
p(z) +
1
\gamma
zp\prime (z) \prec h(z), (1.5)
ISSN 1027-3190. Укр. мат. журн., 2018, т. 70, № 5
SECOND-ORDER DIFFERENTIAL SUBORDINATIONS ON A CLASS . . . 589
then
p(z) \prec q(z) \prec h(z),
where
q(z) =
\gamma
nz\gamma /n
z\int
0
t\gamma /n - 1h(t)dt, z \in \BbbU .
The function q is convex and is the best dominant of the subordination (1.5).
Lemma 1.2 [7]. Let \Re \{ \mu \} > 0, n \in \BbbN and let
w =
n2 + | \mu | 2 -
\bigm| \bigm| n2 - \mu 2
\bigm| \bigm|
4n\Re \{ \mu \}
.
Let h be an analytic function in \BbbU with h(0) = 1 and suppose that
\Re
\Biggl\{
1 +
zh
\prime \prime
(z)
h\prime (z)
\Biggr\}
> - w.
If
p(z) = 1 + pnz
n + pn+1z
n+1 + . . .
is analytic in \BbbU and
p(z) +
1
\mu
zp\prime (z) \prec h(z), (1.6)
then
p(z) \prec q(z),
where q is a solution of the differential equation
q(z) +
n
\mu
zq\prime (z) = h(z), q(0) = 1,
given by
q(z) =
\mu
nz\mu /n
z\int
0
t\mu /n - 1h(t)dt, z \in \BbbU .
Moreover, q is the best dominant of the differential subordination (1.6).
Lemma 1.3 [9]. Let r be a convex function in \BbbU and let
h(z) = r(z) + n\beta zr\prime (z), z \in \BbbU ,
where \beta > 0 and n \in \BbbN . If
p(z) = r(0) + pnz
n + pn+1z
n+1 + . . . , z \in \BbbU ,
ISSN 1027-3190. Укр. мат. журн., 2018, т. 70, № 5
590 A. AKGÜL
is holomorphic in \BbbU and
p(z) + \beta zp\prime (z) \prec h(z), z \in \BbbU ,
then
p(z) \prec r(z),
and this result is sharp.
In the present paper, making use of the subordination results of [5] and [7] we will prove our
main results.
2. Main results.
Theorem 2.1. The set \Re \mu ,\theta (\beta ) is convex.
Proof. Let
fj(z) = z +
\infty \sum
k=2
ak,jz
k, z \in \BbbU , j = 1, . . . ,m,
be in the class \Re \mu ,\theta (\beta ). Then, by the Definition 1.1, we get
\Re
\biggl\{ \Bigl(
R\theta
\mu f(z)
\Bigr) \prime \biggr\}
= \Re
\Biggl\{
1 +
\infty \sum
k=2
L(k, \mu , \theta )ak,jkz
k - 1
\Biggr\}
> \beta . (2.1)
For any positive numbers \lambda 1, \lambda 2, . . . , \lambda m such that
m\sum
j=1
\lambda j = 1.
We have to show that the function
h(z) =
m\sum
j=1
\lambda jfj(z)
member of \Re \mu ,\theta (\beta ); that is,
\Re
\biggl\{ \Bigl(
R\theta
\mu h(z)
\Bigr) \prime \biggr\}
> \beta . (2.2)
Thus, we have
R\theta
\mu h(z) = z +
\infty \sum
k=2
L(k, \mu , \theta )
\left( m\sum
j=1
\lambda jak,j
\right) zk. (2.3)
If we differentiate (2.3) with respect to z, then we obtain
\Bigl(
R\theta
\mu h(z)
\Bigr) \prime
= 1 +
\infty \sum
k=2
kL(k, \mu , \theta )
\left( m\sum
j=1
\lambda jak,j
\right) zk - 1.
ISSN 1027-3190. Укр. мат. журн., 2018, т. 70, № 5
SECOND-ORDER DIFFERENTIAL SUBORDINATIONS ON A CLASS . . . 591
Thus, we have
\Re
\biggl\{ \Bigl(
R\theta
\mu h(z)
\Bigr) \prime \biggr\}
= 1 +
m\sum
j=1
\lambda j\Re
\Biggl\{ \infty \sum
k=2
kL(k, \mu , \theta )ak,jz
k - 1
\Biggr\}
>
> 1 +
m\sum
j=1
\lambda j(\beta - 1) (by (2.2)) = \beta .
Thus, the inequality (2.1) holds and we have desired result.
Theorem 2.1 is proved.
Theorem 2.2. Let q be convex function in \BbbU with q(0) = 1 and let
h(z) = q(z) +
1
\gamma + 1
zq\prime (z), z \in \BbbU ,
where \gamma is a complex number with \Re \{ \gamma \} > - 1. If f \in \Re \mu ,\theta (\beta ) and \digamma = \Upsilon \gamma f, where
\digamma (z) = \Upsilon \gamma f(z) =
\gamma + 1
z\gamma
z\int
0
t\gamma - 1f(t)dt, (2.4)
then \Bigl(
R\theta
\mu f(z)
\Bigr) \prime
\prec h(z) (2.5)
implies \Bigl(
R\theta
\mu \digamma (z)
\Bigr) \prime
\prec q(z),
and this result is sharp.
Proof. From the equality (2.4), we can write
z\gamma \digamma (z) = (\gamma + 1)
z\int
0
t\gamma - 1f(t)dt. (2.6)
By differentiating (2.6) with respect to z, we obtain
(\gamma )\digamma (z) + z\digamma \prime (z) = (\gamma + 1) f(z),
and by applying the operator R\theta
\mu to the last equation, then we get
(\gamma )R\theta
\mu \digamma (z) + z
\Bigl(
R\theta
\mu \digamma (z)
\Bigr) \prime
= (\gamma + 1)R\theta
\mu f(z). (2.7)
If we differentiate (2.7) with respect to z, we have\Bigl(
R\theta
\mu \digamma (z)
\Bigr) \prime
+
1
\gamma + 1
z
\Bigl(
R\theta
\mu \digamma (z)
\Bigr) \prime \prime
=
\Bigl(
R\theta
\mu f(z)
\Bigr) \prime
. (2.8)
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592 A. AKGÜL
By using the differential subordination given by (2.5) in the equality (2.8), we obtain\Bigl(
R\theta
\mu \digamma (z)
\Bigr) \prime
+
1
\gamma + 1
z
\Bigl(
R\theta
\mu \digamma (z)
\Bigr) \prime \prime
\prec h(z). (2.9)
Now, we define
p(z) =
\Bigl(
R\theta
\mu \digamma (z)
\Bigr) \prime
. (2.10)
Then by a simple computation we get
p(z) =
\Biggl[
z +
\infty \sum
k=2
L(k, \mu , \theta )
\gamma + 1
\gamma + k
akz
k
\Biggr] \prime
=
= 1 + p1z + p2z
2 + . . . , p \in H[1, 1].
Using (2.10) in the subordination (2.9), we obtain
p(z) +
1
\gamma + 1
zp\prime (z) \prec h(z) = q(z) +
1
\gamma + 1
zq\prime (z), z \in \BbbU .
If we use Lemma 1.2, we write
p(z) \prec q(z).
So we obtain the desired result and q is the best dominant.
Theorem 2.2 is proved.
Example 2.1. If we choose in Theorem 2.1
\gamma = i+ 1, q(z) =
1 + z
1 - z
,
thus we get
h(z) =
(i+ 2) - ((i+ 2)z + 2)z
(i+ 2)(1 - z)2
.
If f \in \Re \mu ,\theta (\beta ) and \digamma is given by
\digamma (z) = \Upsilon if(z) =
i+ 2
zi+1
z\int
0
tif(t)dt,
then, by Theorem 2.2, we obtain\Bigl(
R\theta
\mu f(z)
\Bigr) \prime
\prec h(z) =
(i+ 2) - ((i+ 2)z + 2)z
(i+ 2)(1 - z)2
=\Rightarrow
=\Rightarrow
\Bigl(
R\theta
\mu \digamma (z)
\Bigr) \prime
\prec 1 + z
1 - z
.
ISSN 1027-3190. Укр. мат. журн., 2018, т. 70, № 5
SECOND-ORDER DIFFERENTIAL SUBORDINATIONS ON A CLASS . . . 593
Theorem 2.3. Let \Re \{ \gamma \} > - 1 and let
w =
1 + | \gamma + 1| 2 -
\bigm| \bigm| \gamma 2 + 2\gamma
\bigm| \bigm|
4\Re \{ \gamma + 1\}
.
Let h be an analytic function in \BbbU with h(0) = 1 and suppose that
\Re
\Biggl\{
1 +
zh
\prime \prime
(z)
h\prime (z)
\Biggr\}
> - w.
If f \in \Re \mu ,\theta (\beta ) and \digamma = \Upsilon \theta
\gamma f, where \digamma is defined by (2.4). Then\Bigl(
R\theta
\mu f(z)
\Bigr) \prime
\prec h(z) (2.11)
implies \Bigl(
R\theta
\mu \digamma (z)
\Bigr) \prime
\prec q(z),
where q is the solution of the differential equation
h(z) = q(z) +
1
\gamma + 1
zq\prime (z), q(0) = 1,
given by
q(z) =
\gamma + 1
z\gamma +1
z\int
0
t\gamma f(t)dt.
Moreover, q is the best dominant of the subordination (2.11).
Proof. If we choose n = 1 and \mu = \gamma + 1 in Lemma 1.2, then the proof is hold by means of
the Theorem 2.2.
Theorem 2.4. Let
h(z) =
1 + (2\beta - 1)z
1 + z
, 0 \leq \beta < 1, (2.12)
be convex in \BbbU , with h(0) = 1 and 0 \leq \beta < 1. If f \in A and verifies the differential subordination\Bigl(
R\theta
\mu f(z)
\Bigr) \prime
\prec h(z),
then \Bigl(
R\theta
\mu \digamma (z)
\Bigr) \prime
\prec q(z) = (2\beta - 1) +
2(1 - \beta )(\gamma + 1)\tau (\gamma )
z\gamma +1
.
Where \tau is given by
\tau (\gamma ) =
z\int
0
t\gamma
t+ 1
dt (2.13)
and \digamma given by equation (2.4). The function q is convex and is the best dominant.
ISSN 1027-3190. Укр. мат. журн., 2018, т. 70, № 5
594 A. AKGÜL
Proof. If
h(z) =
1 + (2\beta - 1)z
1 + z
, 0 \leq \beta < 1,
then h is convex and, by means of Theorem 2.3, we have\Bigl(
R\theta
\mu \digamma (z)
\Bigr) \prime
\prec q(z).
By using Lemma 1.1, we get
q(z) =
\gamma + 1
z\gamma +1
z\int
0
t\gamma h(t)dt =
\gamma + 1
z\gamma +1
z\int
0
t\gamma
\biggl[
1 + (2\beta - 1)t
1 + t
\biggr]
dt =
= (2\beta - 1) +
2(1 - \beta )(\gamma + 1)
z\gamma +1
\tau (\gamma ).
Where \tau is given by (2.13), so we obtain\Bigl(
R\theta
\mu \digamma (z)
\Bigr) \prime
\prec q(z) = (2\beta - 1) +
2(1 - \beta )(\gamma + 1)\tau (\gamma )
z\gamma +1
.
The function q is convex and is the best dominant.
Theorem 2.4 is proved.
Theorem 2.5. If 0 \leq \beta < 1, 0 \leq \mu < 1, \delta \geq 0, \Re \{ \gamma \} > - 1 and \scrF = \Upsilon \gamma f is defined by (2.4),
then we have
\Upsilon \gamma (\Re \mu ,\theta (\beta )) \subset \Re \mu ,\theta (\rho ),
where
\rho = \mathrm{m}\mathrm{i}\mathrm{n}
| z| =1
\Re \{ q(z)\} = \rho (\gamma , \beta ) = (2\beta - 1) + 2(1 - \beta )(\gamma + 1)\tau (\gamma ) (2.14)
and \tau is given by (2.13).
Proof. Let h is given by the equation (2.12), f \in \Re \mu ,\theta (\beta ) and \scrF = \Upsilon \gamma f is defined by (2.4).
Then h is convex and by Theorem 2.3, we deduce\Bigl(
R\theta
\mu \digamma (z)
\Bigr) \prime
\prec q(z) = (2\beta - 1) +
2(1 - \beta )(\gamma + 1)\tau (\gamma )
z\gamma +1
, (2.15)
where \tau is given by (2.13). Since q is convex and q(\BbbU ) is symmetric with respect to the real axis
and \Re \{ \gamma \} > - 1, we have
\Re
\biggl\{ \Bigl(
R\theta
\mu \digamma (z)
\Bigr) \prime \biggr\}
\geq \mathrm{m}\mathrm{i}\mathrm{n}
| z| =1
\Re \{ q(z)\} = \Re \{ q(1)\} = \rho (\gamma , \beta ) =
= (2\beta - 1) + 2(1 - \beta )(\gamma + 1)(1 - \beta )\tau (\gamma ).
From the inequality (2.15), we get
\Upsilon \gamma (\Re \mu ,\theta (\beta )) \subset \Re \mu ,\theta (\rho ),
where \rho is given by (2.14).
Theorem 2.5 is proved.
ISSN 1027-3190. Укр. мат. журн., 2018, т. 70, № 5
SECOND-ORDER DIFFERENTIAL SUBORDINATIONS ON A CLASS . . . 595
Theorem 2.6. Let q be a convex function with q(0) = 1 and h a function such that
h(z) = q(z) + zq\prime (z), z \in \BbbU .
If f \in A, then the subordination
(R\theta
\mu f(z))
\prime \prec h(z) (2.16)
implies that
R\theta
\mu f(z)
z
\prec q(z),
and the result is sharp.
Proof. Let
p(z) =
R\theta
\mu f(z)
z
. (2.17)
Differentiating (2.17), we have
(R\theta
\mu f(z))
\prime = p(z) + zp\prime (z).
If we calculate p(z), then we obtain
p(z) =
R\theta
\mu f(z)
z
=
z +
\sum \infty
k=2
L(k, \mu , \theta )akz
k
z
=
= 1 + p1z + p2z
2 + . . . , p \in H[1, 1]. (2.18)
Using (2.18) in the subordination (2.16) we get
p(z) + zp\prime (z) \prec h(z) = q(z) + zq\prime (z).
Hence by applying Lemma 1.3, we conclude that
p(z) \prec q(z)
that is,
R\theta
\mu f(z)
z
\prec q(z),
and this result is sharp and q is the best dominant.
Theorem 2.6 is proved.
Example 2.2. If we take \mu = 0, \theta = 1 in equality (1.4) and q(z) =
1
1 - z
in Theorem 2.6, then
h(z) =
1
(1 - z)2
and
ISSN 1027-3190. Укр. мат. журн., 2018, т. 70, № 5
596 A. AKGÜL
R1
0f(z) = z +
\infty \sum
k=2
\Gamma (k + 1)akz
k. (2.19)
Differentiating (2.19) with respect to z, we get
\bigl(
R1
0f(z)
\bigr) \prime
= 1 +
\infty \sum
k=2
\Gamma (k + 1)kakz
k - 1 =
= 1 + p1z + p2z
2 + . . . , p \in H[1, 1].
By using Theorem 2.6, we have\bigl(
R1
0f(z)
\bigr) \prime \prec h(z) =
1
(1 - z)2
implies
R1
0f(z)
z
\prec q(z) =
1
1 - z
.
Theorem 2.7. Let
h(z) =
1 + (2\beta - 1)z
1 + z
, z \in \BbbU ,
be convex in \BbbU , with h(0) = 1 and 0 \leq \beta < 1. If f \in A satisfies the differential subordination
(R\theta
\mu f(z))
\prime \prec h(z), (2.20)
then
R\theta
\mu f(z)
z
\prec q(z) = (2\beta - 1) +
2(1 - \beta ) \mathrm{l}\mathrm{n} (1 + z)
z
.
The function q is convex and is the best dominant.
Proof. Let
p(z) =
R\theta
\mu f(z)
z
= 1 + p1z + p2z
2 + . . . , p \in H[1, 1]. (2.21)
Differentiating (2.21), we have
(R\theta
\mu f(z))
\prime = p(z) + zp\prime (z). (2.22)
Using (2.22), the differential subordination (2.20) becomes
(R\theta
\mu f(z))
\prime \prec h(z) =
1 + (2\beta - 1)z
1 + z
.
By using Lemma 1.1, we deduce
p(z) \prec q(z) =
1
z
\int
h(t)dt = (2\beta - 1) +
2(1 - \beta ) \mathrm{l}\mathrm{n} (1 + z)
z
.
Using the relation (2.21) we obtain desired result.
Theorem 2.7 is proved.
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SECOND-ORDER DIFFERENTIAL SUBORDINATIONS ON A CLASS . . . 597
Corollary. If f \in \Re \mu ,\theta (\beta ), then
\Re
\Biggl(
R\theta
\mu f(z)
z
\Biggr)
> (2\beta - 1) + 2(1 - \beta ) \mathrm{l}\mathrm{n} (2) .
Proof. If f \in \Re \mu ,\theta (\beta ), then from Definition 1.1
\Re
\biggl\{ \Bigl(
R\theta
\mu f(z)
\Bigr) \prime \biggr\}
> \beta , z \in \BbbU ,
which is equivalent to
(R\theta
\mu f(z))
\prime \prec h(z) =
1 + (2\beta - 1)z
1 + z
.
Using Theorem 2.7, we have
R\theta
\mu f(z)
z
\prec q(z) = (2\beta - 1) +
2(1 - \beta ) \mathrm{l}\mathrm{n} (1 + z)
z
.
Since q is convex and q (\BbbU ) is symmetric with respect to the real axis, we deduce that
\Re
\Biggl(
R\theta
\mu f(z)
z
\Biggr)
> \Re q(1) = (2\beta - 1) + 2(1 - \beta ) \mathrm{l}\mathrm{n} (2) .
Theorem 2.8. Let q be a convex function such that q(0) = 1 and let h be the function
h(z) = q(z) + zq\prime (z), z \in \BbbU .
If f \in A and verifies the differential subordination\Biggl(
zR\theta
\mu f(z)
R\theta
\mu \digamma (z)
\Biggr) \prime
\prec h(z), z \in \BbbU , (2.23)
then
R\theta
\mu f(z)
R\theta
\mu \digamma (z)
\prec q(z), z \in \BbbU ,
and this result is sharp.
Proof. For the function f \in A, given by the equation (1.1), we have
R\theta
\mu \digamma (z) = z +
\infty \sum
k=2
L(k, \mu , \theta )
\gamma + 1
k + \gamma
akbkz
k, z \in \BbbU .
Let us consider
p(z) =
R\theta
\mu f(z)
R\theta
\mu \digamma (z)
=
z +
\sum \infty
k=2
L(k, \mu , \theta )akbkz
k
z +
\sum \infty
k=2
L(k, \mu , \theta )
\gamma + 1
k + \gamma
akbkz
k
=
ISSN 1027-3190. Укр. мат. журн., 2018, т. 70, № 5
598 A. AKGÜL
=
1 +
\sum \infty
k=2
L(k, \mu , \theta )akbkz
k - 1
1 +
\sum \infty
k=2
L(k, \mu , \theta )
\gamma + 1
k + \gamma
akbkz
k - 1
.
We get
(p(z))\prime =
(R\theta
\mu f(z))
\prime
R\theta
\mu \digamma (z)
- p(z)
(R\theta
\mu \digamma (z))\prime
R\theta
\mu \digamma (z)
.
Then
p(z) + zp\prime (z) =
\Biggl(
zR\theta
\mu f(z)
R\theta
\mu \digamma (z)
\Biggr) \prime
, z \in \BbbU . (2.24)
Using the relation (2.24) in the inequality (2.23), we obtain
p(z) + zp\prime (z) \prec h(z) = q(z) + zq\prime (z)
and, by using Lemma 1.3,
p(z) \prec q(z),
that is,
R\theta
\mu f(z)
R\theta
\mu \digamma (z)
\prec q(z).
Theorem 2.8 is proved.
References
1. Akgül A. On second-order differential subordinations for a class of analytic functions defined by convolution //
Contemp. Anal. and Appl. Math. – 2016.
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4, № 2. – P. 162 – 173.
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Math. – 2012. – № 29. – P. 125 – 129.
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Komatu integral operator // ISRN Math. Anal. – 2014. – № 5. – Art. ID 606235.
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Lect. Notes Math. – 1981. – 1013. – P. 362 – 372.
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Received 11.09.16
ISSN 1027-3190. Укр. мат. журн., 2018, т. 70, № 5
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| id | umjimathkievua-article-1579 |
| institution | Ukrains’kyi Matematychnyi Zhurnal |
| keywords_txt_mv | keywords |
| language | English |
| last_indexed | 2026-03-24T02:08:28Z |
| publishDate | 2018 |
| publisher | Institute of Mathematics, NAS of Ukraine |
| record_format | ojs |
| resource_txt_mv | umjimathkievua/a3/4bbd813a1285ee22e20433230bf461a3.pdf |
| spelling | umjimathkievua-article-15792019-12-05T09:19:33Z Second-order differential subordinations on a class of analytic functions defined by Rafid-operator Диференцiальнi пiдпорядкування другого порядку на класi аналiтичних функцiй, що визначенi оператором Рафiда Akgül, A. Акгюл, А. The purpose of the present paper is to introduce a new class of analytic functions by using the Rafid-integral operator and obtain some subordination results. За допомогою оператора Рафiда введено новий клас аналiтичних функцiй. Отримано деякi результати щодо пiдпорядкування. Institute of Mathematics, NAS of Ukraine 2018-05-25 Article Article application/pdf https://umj.imath.kiev.ua/index.php/umj/article/view/1579 Ukrains’kyi Matematychnyi Zhurnal; Vol. 70 No. 5 (2018); 587-598 Український математичний журнал; Том 70 № 5 (2018); 587-598 1027-3190 en https://umj.imath.kiev.ua/index.php/umj/article/view/1579/561 Copyright (c) 2018 Akgül A. |
| spellingShingle | Akgül, A. Акгюл, А. Second-order differential subordinations on a class of analytic functions defined by Rafid-operator |
| title | Second-order differential subordinations on a class of analytic functions defined
by Rafid-operator |
| title_alt | Диференцiальнi пiдпорядкування другого порядку на класi аналiтичних функцiй, що визначенi оператором Рафiда |
| title_full | Second-order differential subordinations on a class of analytic functions defined
by Rafid-operator |
| title_fullStr | Second-order differential subordinations on a class of analytic functions defined
by Rafid-operator |
| title_full_unstemmed | Second-order differential subordinations on a class of analytic functions defined
by Rafid-operator |
| title_short | Second-order differential subordinations on a class of analytic functions defined
by Rafid-operator |
| title_sort | second-order differential subordinations on a class of analytic functions defined
by rafid-operator |
| url | https://umj.imath.kiev.ua/index.php/umj/article/view/1579 |
| work_keys_str_mv | AT akgula secondorderdifferentialsubordinationsonaclassofanalyticfunctionsdefinedbyrafidoperator AT akgûla secondorderdifferentialsubordinationsonaclassofanalyticfunctionsdefinedbyrafidoperator AT akgula diferencialʹnipidporâdkuvannâdrugogoporâdkunaklasianalitičnihfunkcijŝoviznačenioperatoromrafida AT akgûla diferencialʹnipidporâdkuvannâdrugogoporâdkunaklasianalitičnihfunkcijŝoviznačenioperatoromrafida |